From b83ae140062a924aff7690d7efd05e2ba2185a5c Mon Sep 17 00:00:00 2001
From: Sreeja-99 <75175169+Sreeja-99@users.noreply.github.com>
Date: Sun, 4 Jan 2026 13:12:37 -0600
Subject: [PATCH 1/2] Add multiple solutions for house robber problem

Implemented four different approaches to solve the house robber problem, including recursion, memoization, and dynamic programming.
---
 Leetcode_198.java | 110 ++++++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 110 insertions(+)
 create mode 100644 Leetcode_198.java

diff --git a/Leetcode_198.java b/Leetcode_198.java
new file mode 100644
index 00000000..e0d278bc
--- /dev/null
+++ b/Leetcode_198.java
@@ -0,0 +1,110 @@
+//Way:1
+//Recurssion-- choose or no choose
+//TC: 2^(O(n))
+class Solution {
+    public int rob(int[] nums) {
+        return helper(nums,0);
+        
+    }
+
+    private int helper (int[] nums, int idx){
+        //base case
+        if(idx>=nums.length) return 0;
+
+        //logic
+        //no choose
+        int case1=helper(nums,idx+1);
+        //choose
+        int case2=nums[idx]+helper(nums,idx+2);
+
+        return Math.max(case1,case2);
+
+    }
+}
+
+//Way:2
+//Recurssion-- choose or no choose -- causing time limit exceeded
+//So, going with memoization-- remembering the result
+//TC: O(n)
+class Solution {
+    public int rob(int[] nums) {
+        int[] memo = new int[nums.length];
+        return helper(nums,0,memo);
+        
+    }
+
+    private int helper (int[] nums, int idx, int[] memo){
+        //base case
+        if(idx>=nums.length) return 0;
+
+        if(memo[idx]!=0) return memo[idx];
+
+        //logic
+        //no choose
+        int case1=helper(nums,idx+1,memo);
+        //choose
+        int case2=nums[idx]+helper(nums,idx+2,memo);
+        memo[idx]=Math.max(case1,case2);
+
+        return Math.max(case1,case2);
+
+    }
+}
+
+//Way:3
+//Recurssion-- choose or no choose -- causing time limit exceeded
+//So, going with memoization-- remembering the result
+//TC: O(n)
+class Solution {
+    public int rob(int[] nums) {
+        int[] memo = new int[nums.length];
+        //TO avoid time limit exceeded for the input with all elements as 0
+        for(int i=0;i<nums.length;i++){
+            memo[i]=Integer.MIN_VALUE;
+        }
+        return helper(nums,0,memo);
+        
+    }
+
+    private int helper (int[] nums, int idx, int[] memo){
+        //base case
+        if(idx>=nums.length) return 0;
+
+        if(memo[idx]!=Integer.MIN_VALUE) return memo[idx];
+
+        //logic
+        //no choose
+        int case1=helper(nums,idx+1,memo);
+        //choose
+        int case2=nums[idx]+helper(nums,idx+2,memo);
+        memo[idx]=Math.max(case1,case2);
+
+        return Math.max(case1,case2);
+
+    }
+}
+
+//Way4:
+//DP: tabulation
+//We are choosing alternate houses. We can have one dp array which holds maximum value at that particular index
+//At every index we have two options. To choose the house or not to choose the house
+//Choose the house: sum of curr house and max value at (curr house -2)
+//Not to choose the house: Max value at prev house i.e., currhouse-1
+//Choose max value from above steps
+//TC: O(n); SC: O(n)
+class Solution {
+    public int rob(int[] nums) {
+        int n=nums.length;
+        if(n==1) return nums[0];
+        int[] dp=new int[n];
+        dp[0]=nums[0];
+        dp[1]=Math.max(nums[0],nums[1]);
+
+        for(int i=2;i<n;i++){
+            dp[i]=Math.max(dp[i-1],nums[i]+dp[i-2]);
+        }
+
+        return dp[n-1];
+
+    }
+}

From bafa2bdf8456d1629603ee5c2e13f413f1160f50 Mon Sep 17 00:00:00 2001
From: Sreeja-99 <75175169+Sreeja-99@users.noreply.github.com>
Date: Sun, 4 Jan 2026 13:14:49 -0600
Subject: [PATCH 2/2] Implement multiple approaches for coin change problem

---
 Leetcode_322.java | 137 ++++++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 137 insertions(+)
 create mode 100644 Leetcode_322.java

diff --git a/Leetcode_322.java b/Leetcode_322.java
new file mode 100644
index 00000000..b9ef1156
--- /dev/null
+++ b/Leetcode_322.java
@@ -0,0 +1,137 @@
+//Way1
+//Exhaustive - choose and no choose
+//TC: 2^(m+n); SC: stack space
+class Solution {
+    public int coinChange(int[] coins, int amount) {
+        int ans= helper(coins,amount,0);
+        if(ans==Integer.MAX_VALUE-10) return -1;
+         return ans;
+        
+    }
+
+    private int helper(int[] coins, int amount, int idx){
+        //base case
+        if(idx==coins.length || amount<0) return Integer.MAX_VALUE-10;
+
+        if(amount==0) return 0;
+
+        //logic:
+        //no choose
+        int case1 = helper(coins,amount,idx+1);
+
+        //choose
+        int case2=1+helper(coins,amount-coins[idx],idx);
+
+        return Math.min(case1,case2);
+
+
+    }
+}
+
+
+//Way2:
+//When we have opted for choose and no choose case, we got many repeated sub problems. So, we can choose dp to solve it.
+//In dp, going with bottom up i.e., tabulation method.
+//TC: O(m*n); SC: O(m*n)
+class Solution {
+    public int coinChange(int[] coins, int amount) {
+        int m=coins.length;
+        int n=amount;
+        int[][] dp=new int[m+1][n+1];
+
+        //Number of coins required to make amount 0 with coins 0
+        dp[0][0]=0;
+        //Number of coins required to make amount n with coins 0
+        for(int i=1;i<=n;i++){
+            dp[0][i]=Integer.MAX_VALUE-10;
+        }
+
+        for(int i=1;i<=m;i++){
+            for(int j=0;j<=n;j++){
+                //Using previous rows to get current row.
+                if(j<coins[i-1]){
+                    dp[i][j]=dp[i-1][j];
+                }else{
+                    dp[i][j]=Math.min(dp[i-1][j],1+dp[i][j-coins[i-1]]);
+                }
+                
+            }
+            
+        }
+
+        if(dp[m][n]==Integer.MAX_VALUE-10) return -1;
+        return dp[m][n];
+    }
+}
+
+//Way3:
+//When we have opted for choose and no choose case, we got many repeated sub problems. So, we can choose dp to solve it.
+//In dp, going with bottom up i.e., tabulation method.
+//TC: O(m*n); SC: O(n)
+class Solution {
+    public int coinChange(int[] coins, int amount) {
+        int m=coins.length;
+        int n=amount;
+        int[] dp=new int[n+1];
+
+        //Number of coins required to make amount 0 with coins 0
+        dp[0]=0;
+        //Number of coins required to make amount n with coins 0
+        for(int i=1;i<=n;i++){
+            dp[i]=Integer.MAX_VALUE-10;
+        }
+
+      
+        for(int i=1;i<=m;i++){
+            for(int j=0;j<=n;j++){
+                //Using previous rows to get current row.
+                if(j<coins[i-1]){
+                    dp[j]=dp[j];
+                }else{
+                    dp[j]=Math.min(dp[j],1+dp[j-coins[i-1]]);
+                }
+            }
+        }
+
+        if(dp[n]==Integer.MAX_VALUE-10) return -1;
+        return dp[n];
+    }
+}
+
+//Way4:
+//Exhaustive - choose and no choose -- causing time limit exceeded
+//So, opt for exhaustive with memoization. It means remembering the prev results
+//TC: O(m*n); SC: O(m*n)
+class Solution {
+    public int coinChange(int[] coins, int amount) {
+        int m=coins.length;
+        int n=amount;
+
+        int[][] memo=new int[m+1][n+1];
+
+        int ans= helper(coins,amount,0,memo);
+        if(ans==Integer.MAX_VALUE-10) return -1;
+         return ans;
+        
+    }
+
+    private int helper(int[] coins, int amount, int idx, int[][] memo){
+        //base case
+        if(idx==coins.length || amount<0) return Integer.MAX_VALUE-10;
+
+        if(amount==0) return 0;
+
+        if(memo[idx][amount]!=0) return memo[idx][amount];
+
+        //logic:
+        //no choose
+        int case1 = helper(coins,amount,idx+1,memo);
+
+        //choose
+        int case2=1+helper(coins,amount-coins[idx],idx,memo);
+        memo[idx][amount]=Math.min(case1,case2);
+        return Math.min(case1,case2);
+
+
+    }
+}